I need some help with in trying to prove that $$ \int_{0}^{\infty}{\rm e}^{-sx}\,\frac{\sin^{2}(x)}{x}\,{\rm d}x = \frac{1}{4}\ln\left(1- \frac{4}{s^{2}}\right) \quad\mbox{using}\ Tonelli\mbox{--}Fubini $$ Now, by Tonelli-Fubini we have that : $$ \int_0^{\infty}\int_0^1 e^{-sx}{\sin(2xy)}dydx = \int_0^1\int_0^{\infty}e^{-sx}{\sin(2xy)}dxdy $$ After some calculation, we find that $$ \int_0^{\infty}\int_0^1 e^{-sx}{\sin(2xy)}dydx = \int_0^{\infty}e^{-sx}\frac{\sin(x)^2}{x}dx $$ $$ \mbox{Now we need to prove that}\ \int_0^1\int_0^{\infty}e^{-sx}{\sin(2xy)}dxdy = \frac{1}{4}\ln\left(1- \frac{4}{s^2}\right) $$

This is when I need help : How can I show this equality? I have been thinking but I cannot come up with anything. I will appreciate some help.

  • 1
    $\begingroup$ Do you know the laplace transform of sine? proofwiki.org/wiki/Laplace_Transform_of_Sine $\endgroup$
    – Zima
    Commented 2 days ago
  • 3
    $\begingroup$ applying two times integration by parts, you can solve $\int e^{ax}\sin (bx)dx$ $\endgroup$ Commented 2 days ago
  • 2
    $\begingroup$ I think you mean $\frac14\ln(1\color{blue}{+}\tfrac{4}{s^2})$. $\endgroup$
    – J.G.
    Commented 2 days ago
  • $\begingroup$ Use Euler representation of $\sin(bx)$ $\endgroup$ Commented 2 days ago


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